如何执行日期格式

Question

I want to use the date command to output a day of week from user input.

I want to force the input to be of the format MM/DD/YYYY.

For example, at the command line I give

./programname MM/DD/YYYY MM/DD/YYYY

Snippets from the script itself

#!/bin/bash

DATE_FORMAT="^[0-9][0-9][/][0-9][0-9][/][0-9][0-9][0-9][0-9]$" #MM/DD/YYYY
DATE1="$1"
DATE2="$2"

... followed by

if [ "$DATE1" != "$DATE_FORMAT" ] || [ "$DATE2" != "$DATE_FORMAT" ]; then 
    echo -e "Please follow the valid format MM/DD/YYYY.\n" 1>&2
    exit 1

Now the problem is even when I enter correct date formats,

./programname 11/22/2014 11/23/2014

I still get that error message that I set up, which means that condition for if is evaluated true even when I input valid format... any suggestions why this is happening?

Answer 1

This script seems to work:

#!/bin/bash

DATE_FORMAT="^[01][0-9][/][0-3][0-9][/][0-9][0-9][0-9][0-9]$" #MM/DD/YYYY
DATE1="$1"
DATE2="$2"

if [[ "$DATE1" =~ $DATE_FORMAT ]] && [[ "$DATE2" =~ $DATE_FORMAT ]]
then echo "Both dates ($DATE1 and $DATE2) are OK"
else echo "Please follow the valid format MM/DD/YYYY ($DATE1 or $DATE2 is wrong)."
fi

It uses the =~ operator for a positive regex match inside Bash's [[ test command . The documents don't mention a !~ for negative matching (though that's what Awk and Perl use). With the single-bracket [ test command, there is no regex matching. Note that the regex expression must not be enclosed in double quotes:

Any part of the pattern may be quoted to force the quoted portion to be matched as a string. Bracket expressions in regular expressions must be treated carefully, since normal quoting characters lose their meanings between brackets. If the pattern is stored in a shell variable, quoting the variable expansion forces the entire pattern to be matched as a string.

The test is also more stringent, rejecting 23/45/2091 , amongst other invalid date strings.

$ bash dt19.sh  11/22/2014 11/23/2014
Both dates (11/22/2014 and 11/23/2014) are OK
$ bash dt19.sh  31/22/2014 11/43/2014
Please follow the valid format MM/DD/YYYY (31/22/2014 or 11/43/2014 is wrong).
$

Answer 2

Corrected code:

#!/bin/bash
DATE1="$1"
DATE2="$2"
if echo "$DATE1" | grep -q -E '[0-9][0-9][/][0-9][0-9][/][0-9][0-9][0-9][0-9]'
then 
    echo "Do whatever you want here"
    exit 1
else
    echo "Invalid date"
fi